Answer:
Quadrant I: (1,1), (4,3)
Quadrant II: (-2, 3), (-1, 1)
Step-by-step explanation:
Quadrant I points have positive x and y values. Quandrant II points have negative x values and positive y values.
Answer: t-half = ln(2) / λ ≈ 0.693 / λExplanation:The question is incomplete, so I did some research and found the complete question in internet.
The complete question is:
Suppose a radioactive sample initially contains
N0unstable nuclei. These nuclei will decay into stable
nuclei, and as they do, the number of unstable nuclei that remain,
N(t), will decrease with time. Although there is
no way for us to predict exactly when any one nucleus will decay,
we can write down an expression for the total number of unstable
nuclei that remain after a time t:
N(t)=No e−λt,
where λ is known as the decay constant. Note
that at t=0, N(t)=No, the
original number of unstable nuclei. N(t)
decreases exponentially with time, and as t approaches
infinity, the number of unstable nuclei that remain approaches
zero.
Part (A) Since at t=0,
N(t)=No, and at t=∞,
N(t)=0, there must be some time between zero and
infinity at which exactly half of the original number of nuclei
remain. Find an expression for this time, t half.
Express your answer in terms of N0 and/or
λ.
Answer:
1) Equation given:
← I used α instead of λ just for editing facility..
Where No is the initial number of nuclei.
2) Half of the initial number of nuclei:
N (t-half) = No / 2So, replace in the given equation:
3) Solving for α (remember α is λ)
αt ≈ 0.693
⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ
The property that is applied in
is addition property of equality
Given :

From the above equation we can see that 4 is added at the both sides of the equation .
we can add same number at the both sides of the equation to balance the equation . That is called as addition property of equality
Here , 4 is added on both sides of the equality to balance the equation .
So , addition property of equality is applied .
Learn more : brainly.com/question/24214650
10. Find which one of the given fractions are value closer to 1.
First = 7/8
Second = 11/10
Let’s start solving
Let us start with the first value which is 7/8
=> 7 needs to be divided with 8
=> 7 / 8 = 0.875
Now, let’s solve the second value which is 11/10
=> same with first solution., divide 11 by 10
=> 11 / 10
=> 1.1
Thus, 11/10 is the has the nearest value to 1.